Question

The velocity of a particle ts after it starts from rest is vm/s,where v=1.25t-0.05t^2.Find
i)the initial accelaration of the partcle
ii)the displacement of the partcle from its staring point at the instant when its accelaration is 0.05m/s^2

Answer 1

acceleration = (v-u) /t
= [(1.25t - 0.05t^2) -0] / t
= t(1.25 - 0.05t) /t
= 1.25 - 0.05t ms^-2

Answer 2

ok

Answer 3

when a = 0.05
from (1)
a=1.25 - 0.05t
0.05 = 1.25-0.05t
-1.2 = -0.05t
t = 24s
By s= ut +(1/2)at^2
s = 0(24) + (1/2)(0.05)(24)^2
= do it yourself
Honestly, i'm not sure if i have done (ii) correctly. I just guess it's like that... Sorry :(

Answer 4

i think for the first the initial acceleration should be
1.25- 0.1t by doing differentitation of velocity u ll get acceleration

Answer 5

Trust heena, don't trust me (i'm serious)

Answer 6

nah it happens wid everyone @callissto but how can we find displacement now?

Answer 7

i did it like you calistro and i was wrong.Heena is right.

Answer 8

now at this stage t=0.because its at its initial accelaration.therefore a=1.25m/s^2

Answer 9

By putting a= 0.05 to the equation to find out t,
and then use integration
∫v dt from t=0 to t= (you found in acc.)

Answer 10

v^2-U^2=2as
(1.25t-0.05t^2)^2=2x0.05S
(1.25t-0.05t^2)^2/0.01=s

Answer 11

Once again , don't trust me :(

Answer 12

hey @callisto it happens wid all even wid me too several but the bests thing atleast we tried ok..

Answer 13

Sorry , just wanna ask how do you know the velocity 'at the instant when its accelaration is 0.05m/s^2' is =(1.25t-0.05t^2) ?

Answer 14

well i used the formula
V^2-u^2=2as so i guess we have to do like this but abu t i m still not sure.. :?

Answer 15

one request from u plz kindly post the qn wid der respective group <3

Answer 16

nope, i was asking how do you know at the instant a = 0.05, v is equal to 1.25t-0.05t^2?

Answer 17

oh yess i m a big fool we have to do doubleintegration of acceleration to get displacement :P

Answer 18

its because this partcle is the same particle as for question 1 and also the only thing that has changed is accelaration

Answer 19

when acceleration changes, shouldn't the velocity also change?

Answer 20

S=0.025 t^2
do integration of acceleartion two time to get displacement thnx @callisto :)

Answer 21

One more question, can i do integration once of velocity to get displacement?

Answer 22

Note: it is a definite integral and you should find the respect time for it

Answer 23

No we are given constants on the v equation. yeah hennna.first weneed to find time on the first intergrated formula which leads to acclaration.(a=1.25-0.1t)a is 0.05 therefore t=12.Now we use limints for t as 12 and 0.to get 61.2m

Answer 24

calistro ya u can do that

Answer 25

right, i surrender :(

Answer 26

i dunno abu limits and all dat but i m ssure as in first we are said abu diferentiation and whenever we are said about instant we have to do integration i just cross chck my notes

Answer 27

if v = constant, a = 0, that's what i know :(

Answer 28

yes thats true and whenever u intergrate v=1.25t-0.05t^2 U intergrate with respect to t.For u to get a real number answer n its correct

Answer 29

agree wid CAllisto abu a=0

Answer 30

thanks for giving me a lesson :)

Answer 31

@callisto there is a file for u hope it helps

Answer 32

yup i know that , but still thanks :) i was think if i do integration of velocity instead of 2twice of accelertation, i can reduce the chance of making mistakes

Answer 33

yup u can :)

Answer 34

yep thanks guys

Answer 35

@a level student plz post this qn on physc sec nxt time plz .....

Answer 36

yeah ill do that